Call exec inside a function - NameError-CodePudding

Why does this code raise NameError?:

def func():
    exec("my_var = 42")
    print(my_var)
func()

There are many related questions, but I have not found a clear answer as to WHY. I don't want a workaround.

Also, I have noticed that the code works when I run:

exec("my_var = 42", globals(), globals())

But not really sure why.

The documentation for exec() states: "...if the optional parts are omitted [second and third arguments], the code is executed in the current scope.". The current scope is the func() function. Why can't I access my_var from this same scope?

CodePudding user response：

The parser/compiler doesn't take the argument to exec() into account (since it could be a variable). So when it parses the function it doesn't see the assignment to my_var, so it treats it as a global variable. But the assignment in exec() will create a local variable. So the variable that was assigned is not the same one that it tries to print.

CodePudding user response：

To understand what's happening with this code, you need to try the following situation:

my_var = None
def func():
    exec("my_var = 42")
    print(my_var)
func()

The output is None because the variable is declared as a global variable, outside the function. But if you declare it inside and execute exec("my_var = 42", globals(), globals()) it becomes automatically a global variable that you can use outside your function, as you can see when running globals():

def func():
    exec("my_var = 42", globals(), globals())
    print(my_var)
func()
print(my_var, globals())

Output:

42 {'my_var': 42 ......}